Q:How does one calculate the piezometric head (or flow potential) for a column of gas at rest?

Info: From hydrostatics we know that for a liquid continuum at rest or in uniform flow in a gravity field, pressure variations obey: $$\tag{1} -\frac{\partial p}{\partial z}=\rho g=\gamma; \ \ \frac{\partial p}{\partial x}=\frac{\partial p}{\partial y}=0$$

Hence, for a liquid column of constant density ($\rho$), Eqn 1 can be integrated from two points of elevation in the column to obtain: $$\tag{2} z+p/\gamma=\text{const.} \ \ \text{or:} \ \ p_2=p_1-\rho g(z_2 - z_1)$$

which is the equation for hydrostatic pressure distribution. Using the properties of water, I have calculated the hydrostatic pressure distribution assuming the gauge pressure $p_2 = 0$ Pa at a free surface of elevation $z_2=10$ m to an elevation of $z_1=0$ m. The results of which are shown in the figure below, on the left. enter image description here

The quotient $p/\gamma$ in Eqn 2 is called the pressure head and $z$ represents the elevation head, the sum of which is called the piezometric head $\varphi$ (or Hydraulic head), expressed in units of length. For a fluid column at rest the piezometric head is constant everywhere, as shown plotted red in the figure above on the right, using the same water pressure data (given in the figure below). These figures show the components of the piezometric head $\varphi$ or flow potential $\Phi$ (plot on the left) (Note: there are several ways to define a potential. The flow potential, $\Phi$, used here is in units of pressure rather than length).

enter image description here

I then considered a static column of nitrogen gas. I used the barometric formula to calculate pressure and density as a function of elevation to try to reproduce the results I found assuming a liquid of constant density, i.e., to calculate a constant piezometric head $\varphi$ or flow potential $\Phi$ for the static column of nitrogen gas.

This time, instead of calculating the static pressure distribution from the top of the fluid column, I started from the base ($z_1=0$ m), assuming an absolute pressure of $p_1=101,325$ Pa, and isothermal temperature of 68F (293.15 K), using the barometric formula of the following form:

$$\tag{3} p_2 = p_1 \cdot \exp\left(-\frac{Mg}{RT}(z_2 - z_1)\right)$$

Likewise, for density I used

$$\tag{4} \rho_2 = \rho_1 \cdot \exp\left(-\frac{Mg}{RT}(z_2 - z_1)\right)$$

where the base density ($\rho_1$) used was 1.165 kg/m3. The results of these calculations, shown below, did not produce a constant piezometric head $\varphi$ or flow potential $\Phi$. I think my error stems from how I am calculating the elevation potential, but I am unsure.

enter image description here enter image description here


The issue I was having with the elevation potential, the $\rho g z$ term, was that I was calculating the density of gas at an elevation and then using that (constant) density value for the entire column height. However as we know for gases, density varies with pressure. Therefore, to evaluate the aerostatic/hydrostatic pressure loss/gain, the column of gas should be subdivided into a sufficient number of segments, such that the density in each segment can be assumed to be constant. A numerical integration scheme can be used, such as trapezoidal integration, to sum the contributions of pressure of each subdivision. Alternatively, we can substitute the barometric formula for density, Eqn 4, into our relationship between pressure, density, and elevation, Eqn 1: $\partial p =- \rho g\ \partial z$ and then integrate this equation with respect to elevation, e.g.,

$$\tag{5} \int_{p_1}^{p(z)} dp=g\rho_1\int_{z_1}^{z}{e^{-a(z-z_1)}}\ dz$$

with $a=Mg/RT$. This integrates to

$$\tag{6} p(z)-p_1=\frac{g\rho_1}{a}(1-e^{-a(z-z_1)})$$

See here for details on this integration. Using Equation 6 to calculate the elevation potential produces the expected results as seen in the figure below.

enter image description here

However, I am not quite sure how to calculate the pressure head, $p/\rho g$ or $p/\gamma$....thoughts?

Update 2

For compressible fluid under isothermal conditions the pressure head ($\Psi$) is defined by $$\tag{7} \Psi=\int_{p_1}^p \frac{dp}{g \rho(p)}$$

Assuming an ideal gas, the relationship between pressure and density is $$\tag{8} \rho = \frac{pM}{RT}$$

Substituting Eqn 8 into Eqn 7, and then performing the integration we have $$\tag{9} \ln(p(z)/p_1)/a$$

where $a=\frac{Mg}{RT}$

Using Eqn 9 to calculate the pressure head (as a function of pressure, which is a function of elevation) produced the results seen in the figure below. The hydrostatic or piezometric head remains constant, which indicates that the column of gas is static, a result I expect. However, I'm not sure I feel OK with the negative pressure head values. Thoughts?

enter image description here


closed as unclear what you're asking by Jon Custer, heather, Steeven, Mark Mitchison, Rococo Feb 16 '17 at 6:04

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  • $\begingroup$ I honestly have no idea where you are going with this, and what the point of it all is supposed to be. All I can see in the above is a number of trivial rearrangements of terms, amounting to demonstrating a result of the form $g(z)=C+f(z)-f(z)=C$, with $g$ and $f$ various functions and some constant $C$. This is a thoroughly unsurprising result. And, no, a result of that kind indicates exactly nothing other than that you have done your algebra correctly. What is it that you are after here? $\endgroup$ – Pirx Feb 4 '17 at 17:07
  • $\begingroup$ P.S.: Oh, and of course you get a negative "pressure head" because the pressure drops as you increase altitude. $\endgroup$ – Pirx Feb 4 '17 at 17:18

Assuming an ideal gas and constant temperature as you do the result is trivial: You get


with $R$ the specific gas constant and $T$ the absolute temperature. This is the same result, by the way, which you would obtain by simply combining your equations (3) and (4) above.

I will say that I have never seen anyone using this terminology (piezometric, or hydraulic, head) in the atmospheric sciences. I can't see this concept as being practically useful in this area, but I may be wrong.

It is also worth noting that the assumption of constant temperature doesn't usually make much sense for atmospheric pressure calculations. Standard atmospheric temperature and pressure distributions are discussed here.

  • $\begingroup$ I think for compressible fluid under isothermal conditions the pressure head is defined by: $$\int_{p_o}^p \frac{dp}{g \rho(p)}$$ At this moment, I'm not quite sure how to set this up for spreadsheet calculation. As you can see in my last figure, I was able to calculate a constant Flow Potential using the concepts of pressure and elevation potential. I should be able to do the same in the context of "heads": piezometric, pressure, and elevation. $\endgroup$ – Armadillo Feb 2 '17 at 18:28
  • $\begingroup$ The result of the integration is simply your equation (3), solved for $\Delta z$. $\endgroup$ – Pirx Feb 2 '17 at 18:33
  • $\begingroup$ P.S.: I am not sure what the point of your trivial rearrangements of terms is supposed to be in general, and specifically why you are interested in your "flow potential" $\Phi$. As far as I can tell you have defined this as $\Phi=p_0+\int \rho(h)\,g\,h\,\mbox{d}h-\int \rho(h)\,g\,h\,\mbox{d}h=p_0$. What in all the world is this supposed to be good for? $\endgroup$ – Pirx Feb 2 '17 at 18:46
  • $\begingroup$ Darcy originally expressed fluid flow in terms of "head" (echo2.epfl.ch/VICAIRE/mod_3/chapt_5/main.htm), measured as elevations in manometers (or piezometers). If I am using pressure gauges rather than manometers I should use the rearrangement that allows me to use pressure and calculated elevation potential to calculate the flow potential. Usually, when flowing gas the "elevation head or potential" is neglected. I am trying to solidify my understanding and quantify how much a difference this value is when testing at say 10,000 psi and elevation change of 10 meters. $\endgroup$ – Armadillo Feb 2 '17 at 18:58
  • $\begingroup$ 10 meters at a base pressure of $10{,}000\,\mbox{psi}$ (!!!)? The gravity-induced pressure change in this case is of the order of $100\,\mbox{Pa}$, or $0.014\,\mbox{psi}$, six orders of magnitude less than your base pressure. I doubt you'll have pressure gauges that could tell the difference. There's a reason why atmospheric pressure changes are neglected at such minuscule elevation differences. $\endgroup$ – Pirx Feb 2 '17 at 19:03

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